2.2. Equivalent parameter calculation method of ball screw feed system lumped mass model

Accurate ball screw feed system dynamic model requires a reasonable equivalent parameter calculation method of the lumped mass model. As mentioned the shaft has influence on the rotational mode and the axial mode of the drive system; the shaft is separated into two different branches, an axial branch and a rotational branch, while the coupling is realized using constrained equations. The dynamic characteristics of the feed system should be analyzed to select the appropriate equivalent parameter calculation method. The inertia of the axial system and the inertia of rotational system are not only the mass or inertia of the component itself but also the mass or inertia converted to the independent coordinate system component of the dynamic system.

The equivalent rotary inertia of the screw JS is composed of the rotary inertia of the screw Jsc, the rotary inertia of the coupling Jc, and the mass of the table MT converted to the rotary inertia of the screw:

$$J\_S = J\_{sc} + J\_c + M\_T \dot{\mathbf{r}}^2 \tag{9}$$

screw feed system consists of screw bearing rigidity kb, and the axial rigidity of the screw shaft

With the material elastic modulus E, screw cross-sectional area A, and ball screw length at table position ln, the axial rigidity of the screw according to the position of the table k<sup>s</sup>ax can be

The torsional rigidity of the ball screw feed system krot consists of the torsional rigidity of the screw ksrot and the torsional rigidity of the coupling kc. According to the position of the table, the torsional rigidity of the screw ksrot can be approximated with the shear modulus G and

<sup>¼</sup> <sup>π</sup>d<sup>2</sup> sE 4ln

<sup>¼</sup> <sup>G</sup> � <sup>π</sup> � <sup>d</sup><sup>4</sup>

s 32 � ln

Electromechanical Co-Simulation for Ball Screw Feed Drive System

http://dx.doi.org/10.5772/intechopen.80716

(13)

45

(14)

(15)

(16)

(17)

kax <sup>¼</sup> <sup>1</sup> kb þ 1 ksax �<sup>1</sup>

ksax <sup>¼</sup> EA ln

krot <sup>¼</sup> <sup>1</sup> kc þ 1 ksrot -1

Δθ <sup>¼</sup> <sup>G</sup> � IP ln

With the nut reference rigidity K, nut axial load Fa, and basic dynamic load Ca, the contact

Permanent magnet synchronous motors can be divided into two types according to the rotor type, salient pole rotor and non-salient pole rotor. The structure is shown in Figure 5; in the surface-mounted permanent magnet synchronous motor (Figure 5(a)), the magnetic circuit of the rotor is symmetrical, and the magnetic permeability and air gap permeability of the permanent magnet material are approximately the same. In the rotor two-phase coordinate system, the direct-axis inductance and the quadrature-axis inductance are equal, that is

The rotor magnetic paths of plug-in-type (Figure 5(b)) and built-in-type (Figure 5(c)) permanent magnet synchronous motors are asymmetrical, and the quadrature-axis inductance is

Ld ¼ Lq. It is named non-salient pole rotor permanent magnet synchronous motor.

0:1Ca <sup>1</sup> 3

kn <sup>¼</sup> <sup>0</sup>:8<sup>K</sup> Fa

ksrot <sup>¼</sup> <sup>T</sup>

polar rotary inertia of the screw section IP as shown in Eq. (16):

rigidity of the screw nut kn can be expressed as follows:

3. Servo control system modeling and simulation

3.1. Modeling of permanent magnet synchronous motor

ksax is as follows:

established as shown in Eq. (14):

With the material density r, namely, equivalent diameter ds and length ls of the screw shaft, the rotary inertia of the screw Jsc can be approximated using the following equation:

$$J\_{sc} = \frac{\rho \cdot \pi}{32} d\_s^4 l\_s$$

The screw equivalent mass MS is composed of screw mass Msc, servomotor rotor mass Mm, and coupling mass Mc; with the material density r, namely, equivalent diameter ds and length ls of the screw shaft, the rotary inertia of the screw values Jsc can be approximated using the following equations:

$$M\_S = M\_{sc} + M\_m + M\_c \tag{11}$$

$$M\_{\rm sc} = \frac{\rho \pi}{4} d\_s^2 l\_s \tag{12}$$

The axial rigidity of the ball screw feed system is related to the installation method of the screw. Here is an example of the screw-fixed-support method used on most machine tools (Figure 4). Servomotor side of the screw shaft uses fixed support to provide screw axial support, and the end of the shaft is free support. Therefore, the axial rigidity kax of the ball

Figure 4. Screw fixation-support installation diagram.

screw feed system consists of screw bearing rigidity kb, and the axial rigidity of the screw shaft ksax is as follows:

2.2. Equivalent parameter calculation method of ball screw feed system

Accurate ball screw feed system dynamic model requires a reasonable equivalent parameter calculation method of the lumped mass model. As mentioned the shaft has influence on the rotational mode and the axial mode of the drive system; the shaft is separated into two different branches, an axial branch and a rotational branch, while the coupling is realized using constrained equations. The dynamic characteristics of the feed system should be analyzed to select the appropriate equivalent parameter calculation method. The inertia of the axial system and the inertia of rotational system are not only the mass or inertia of the component itself but also the mass or inertia converted to the independent coordinate system

The equivalent rotary inertia of the screw JS is composed of the rotary inertia of the screw Jsc, the rotary inertia of the coupling Jc, and the mass of the table MT converted to the rotary inertia

JS ¼ Jsc þ Jc þ MTi

With the material density r, namely, equivalent diameter ds and length ls of the screw shaft, the

The screw equivalent mass MS is composed of screw mass Msc, servomotor rotor mass Mm, and coupling mass Mc; with the material density r, namely, equivalent diameter ds and length ls of the screw shaft, the rotary inertia of the screw values Jsc can be approximated using the

Jsc <sup>¼</sup> <sup>r</sup> � <sup>π</sup> <sup>32</sup> <sup>d</sup><sup>4</sup>

Msc <sup>¼</sup> <sup>r</sup><sup>π</sup>

The axial rigidity of the ball screw feed system is related to the installation method of the screw. Here is an example of the screw-fixed-support method used on most machine tools (Figure 4). Servomotor side of the screw shaft uses fixed support to provide screw axial support, and the end of the shaft is free support. Therefore, the axial rigidity kax of the ball

4 d2

rotary inertia of the screw Jsc can be approximated using the following equation:

<sup>2</sup> (9)

<sup>s</sup> ls (10)

<sup>s</sup> ls (12)

MS ¼ Msc þ Mm þ Mc (11)

lumped mass model

44 New Trends in Industrial Automation

of the screw:

following equations:

Figure 4. Screw fixation-support installation diagram.

component of the dynamic system.

$$k\_{\rm ax} = \left(\frac{1}{k\_b} + \frac{1}{k\_{\rm sx}}\right)^{-1} \tag{13}$$

With the material elastic modulus E, screw cross-sectional area A, and ball screw length at table position ln, the axial rigidity of the screw according to the position of the table k<sup>s</sup>ax can be established as shown in Eq. (14):

$$k\_{s\_{\text{ax}}} = \frac{EA}{l\_n} = \frac{\pi d\_s^2 E}{4l\_n} \tag{14}$$

The torsional rigidity of the ball screw feed system krot consists of the torsional rigidity of the screw ksrot and the torsional rigidity of the coupling kc. According to the position of the table, the torsional rigidity of the screw ksrot can be approximated with the shear modulus G and polar rotary inertia of the screw section IP as shown in Eq. (16):

$$k\_{\rm rot} = \left(\frac{1}{k\_{\rm c}} + \frac{1}{k\_{\rm s\_{\rm rot}}}\right)^{-1} \tag{15}$$

$$k\_{s\_{mt}} = \frac{T}{\Delta\theta} = \frac{G \cdot I\_P}{l\_n} = \frac{G \cdot \pi \cdot d\_s^4}{32 \cdot l\_n} \tag{16}$$

With the nut reference rigidity K, nut axial load Fa, and basic dynamic load Ca, the contact rigidity of the screw nut kn can be expressed as follows:

$$k\_n = 0.8K \left(\frac{F\_a}{0.1C\_a}\right)^{\frac{1}{5}} \tag{17}$$
